# Metric Vector Spaces: The Theory of Bilinear Forms

**Book**: Advanced Linear Algebra
**Author**: Steven Roman

###### tags: `study_note` `math` `linear_algebra`

[TOC]

Vector spaces over **arbitrary fields** that have **bilinear form** defined upon them

## Bilinear Forms

[[Wiki: Multilinear Form](https://www.wikiwand.com/en/Multilinear_form)] [[Wiki: Bilinear Form](https://www.wikiwand.com/en/Bilinear_form)]

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**Definition** Let $V$ be a vector space over $F$. A mapping $\langle,\rangle: V\times V\rightarrow F$ is called a **bilinear form** if it is a linear function of each coordinate, that is, if $$\langle\alpha x + \beta y, z\rangle = \alpha \langle x, z\rangle + \beta \langle y, z\rangle$$

and $$\langle z, \alpha x + \beta y\rangle = \alpha \langle z, x\rangle + \beta \langle z, y\rangle$$
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## Symmetric, Skew-Symmetric and Alternate Forms

[[Wiki: Inner Product Space](https://www.wikiwand.com/en/Inner_product_space)] [[Wiki: Metric Space](https://www.wikiwand.com/en/Metric_space)]

A bilinear form is

1. **symmetric** if $$\langle x, y\rangle = \langle y, x\rangle, \forall x, y\in V$$
2. **skew-symmetric** (or **antisymmetric**) if $$\langle x, y\rangle = -\langle y, x\rangle, \forall x, y\in V$$
3. **alternate** (or **alternating**) if $$\langle x, x\rangle=0, \forall x\in V$$

These three kinds of bilinear forms are referred to as an **inner product** and a pair $(V, \langle,\rangle)$ is called a [metric vector space](https://www.wikiwand.com/en/Metric_space) or [inner product space](https://www.wikiwand.com/en/Inner_product_space).

- **Orthogonal geometry** (也稱歐式幾何) over $F$, if $\langle,\rangle$ is *symmetric*.
- **Symplectic geometry** (扭對稱幾何) over $F$, if $\langle,\rangle$ is *alternate*.

**Example [Minkowski Space](https://www.wikiwand.com/en/Minkowski_space)** $M_4$ is the four-dimensional real **orthogonal geometry** $\mathbb{R}^4$ with inner product defined by

- $\langle e_1,e_1\rangle = \langle e_2,e_2\rangle = \langle e_3,e_3\rangle = 1$
- $\langle e_4,e_4\rangle = -1$
- $\langle e_i,e_j\rangle = 0, \forall i\neq j$

where $e_1, \ldots, e_4$ is the standard basis for $\mathbb{R}^4$.

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**Theorem** Let $V$ be a vector space over a field $F$

1. If $\text{char}(F)=2$, then
$$\text{symmetric} \Leftrightarrow \text{skew-symmetric}$$ $$\text{alternate} \Rightarrow \text{skew-symmetric}$$
2. If $\text{char}(F)\neq2$ (ex. $\mathbb{R}, \mathbb{C}$), then $$\text{alternate} \Leftrightarrow \text{skew-symmetric}$$
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## The Matrix of a Bilinear Forms

> Binear form 要變成類似 inner product 的形式
> $$\langle x, y\rangle = x^{\top}y$$必須經由一個 matrix $M_{\mathcal{B}}$ 使得 $$\langle x, y\rangle = [x]_{\mathcal{B}}^{\top}M_{\mathcal{B}}[y]_{\mathcal{B}}$$

If $\mathcal{B}=(b_1, \ldots, b_n)$ is an ordered basis for a metric vector space $V$ then $\langle, \rangle$ is completely determined by the $n\times n$ matrix of values $$M_{\mathcal{B}} = (a_{ij}) = (\langle b_i, b_j\rangle)$$
This is referred to as the matrix of the form $\langle, \rangle$ with respect to the ordered basis $\mathcal{B}$

For all $x, y\in V$, $x, y$ can be represented as $x=\sum_i r_ib_i, y=\sum_j s_jb_j$, then $$\langle x, y\rangle=\langle\sum_i r_ib_i, \sum_j s_jb_j\rangle$$ $$=\sum_i\sum_jr_i\langle b_i, b_j\rangle s_j$$ $$=[r_i]^{\top}M_{\mathcal{B}}[s_j] = [x]_{\mathcal{B}}^{\top}M_{\mathcal{B}}[y]_{\mathcal{B}}$$

- $\langle, \rangle$ is symmetric $\Leftrightarrow M_{\mathcal{B}}$ is symmetric.
- $\langle, \rangle$ is skew-symmetric $\Leftrightarrow M_{\mathcal{B}}$ is skew-symmetric.
- $\langle, \rangle$ is alternate $\Leftrightarrow M_{\mathcal{B}}$ is skew-symmetric and has 0's on the main diagnal.

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**Definition** Two matrices $A, B\in \mathcal{M}_n(F)$ are said to be **congruent** if there exists an invertible matrix $P$ for which $$A=P^{\top}BP$$ The equivalence classes under congruence are called **congruence classes**.
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**Theorem** $\forall A, B\in \mathcal{M}_n(F), A, B$ represent the same bilinear form on $V \Leftrightarrow A, B$ are congruent.
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**Definition** The **discriminant** of a bilinear form is defined as $$\{det(A) | A\in \mathcal{M}_n(F) \text{ represent the bilinear form}\}$$
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If $A$ and $B$ are congruent matrices then $$det(A)=det(P^{\top}BP)=det(P)^2det(B)$$ so $det(A)$ and $det(B)$ differ by a square factor.

## Quadratic Forms

[[Wiki: Quadratic Forms](https://www.wikiwand.com/en/Quadratic_form)]

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**Definition** A **quadratic form** on a vector space $V$ is a map $Q:V\rightarrow F$ with following properties:

1. For all $r\in F, v\in V$ $$Q(rv)=r^2Q(v)$$
2. The map with respect to $Q$ $$\langle u, v \rangle_Q = Q(u+v) - Q(u) - Q(v)$$ is a (symmetric) bilinear form.
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**Example** Let $Q(x)=x^2$

If $\text{char}(F)\neq2$ and $\langle,\rangle$ is symmetric, then we can define $$Q(x)=\frac{1}{2}\langle x, x\rangle$$ such that $$\langle u, v\rangle_Q = \langle u, v\rangle$$

知道了 quadratic form 等同於知道了 bilinear form。

## Orthogonality
[[Wiki: Degenerate Bilinear Forms](https://www.wikiwand.com/en/Degenerate_bilinear_form)]

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**Definition**
- A vector $x$ is **orthogonal** to a vector $y$, written $x\perp y$, if $\langle x, y\rangle=0$
- A vector $x\in V$ is **orthogonal** to a subset $S$ of $V$, written $x\perp S$, if $\langle x, s\rangle=0$, for all $s\in S$.
- A subset $S$ of $V$ is **orthogonal** to a subset $T$ of $V$, written $S\perp T$, if $\langle s, t\rangle=0$, for all $s\in S, t\in T$.
- The **orthogonal complement** of a subset $X$ of a metric vector space $V$, denoted $X^{\perp}$, is the subspace $$X^{\perp}=\{v\in V | v\perp X\}$$
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Regardless of whether the form $\langle,\rangle$ is symmetric or alternate (and hence skew-symmetric), $x\perp y$ always implies $y\perp x$:

- (Symmetric) $\langle x, y\rangle=0\Rightarrow \langle y, x\rangle=0$
- (Alternate) $\langle x, y\rangle=0$

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**Definition** Let $(V, \langle,\rangle)$ be a metric vector space.

1. A nonzero $x\in V$ is **isotropic** (or **null**) if $\langle x, x\rangle=0$; o.w. it is **nonisotropic**.
2. $V$ is **nonisotropic** (also called **anisotropic**) if it contains no isotropic vectors.
3. $V$ is **isotropic** if it contains at least one isotropic vector.
4. $V$ is **totally isotropic** (that is, symplectic) if all vector in $V$ are isotropic.
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**Definition** Let $(V, \langle,\rangle)$ be a metric vector space.
1. The set $V^{\perp}$ of all degenerate vectors is called the **radical** of $V$ and written $$\text{rad}(V)=V^{\perp}=\{x\in V | x\perp v, \forall v \in V\}$$
2. $V$ is **nonsingular**, or **nondegenerate**, if $\text{rad}(V)=\{0\}$.
3. $V$ is **singular**, or **degenerate**, if $\text{rad}(V)\neq \{0\}$.
4. $V$ is **totally singular** or **totally degenerate**, if $\text{rad}(V)=V$.
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If $S$ is a subspace of $V$, then $$\text{rad}(S)=S\cap\text{rad}(V)$$

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**Theorem** Let $\langle, \rangle$ be a bilinear form on $V$. Then orthogonality is a symmetric relation, that is, $$x\perp y\Rightarrow y\perp x$$ if and only if $\langle, \rangle$ is either symmetric or alternate, that is, if and only if $(V, \langle, \rangle)$ is a metric vector space.
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**Proof**
- ($\Leftarrow$)
- Suppose $\langle, \rangle$ is symmetric, $$x\perp y\Rightarrow\langle x, y\rangle=0=\langle y, x\rangle\Rightarrow y\perp x$$
- Suppose $\langle, \rangle$ is alternate (also implies skew-symmetric), $$x\perp y\Rightarrow\langle x, y\rangle=0 = -0 =\langle y, x\rangle\Rightarrow y\perp x$$
- ($\Rightarrow$) Suppose $$x\perp y\Rightarrow y\perp x$$ ...

## Linear Functionals
[[Wiki: Riesz Space](https://www.wikiwand.com/en/Riesz_space)]

Let $V$ be a metric vector space over $F$. Let $x\in V$ and consider the **inner product on the right** map $\phi_x:V\rightarrow F$ defined by $$\phi_x(v)=\langle v,x\rangle$$ This is easily seen to be a linear functional and so we can define a function $\tau:V\rightarrow V^*$ which maps $V$ into its [**dual space**](/s/S1MZuuDyr#Dual-Bases) $V^*$ by $$\tau(x)=\phi_x$$ The bilinearity of the form insures that $\tau$ is linear and the kernel of $\tau$ is $$\text{ker}(\tau)=\{x\in V|\phi_x=0\}=\{x\in V|\langle v,x \rangle=0 ,\forall v\in V\}=V^{\perp}$$

Hence, if $V$ is nonsingular then $\text{ker}(\tau)=V^{\perp}=\{0\}$ and so $\tau$ is injective (one-to-one). Moreover, since $\text{dim}(V)=\text{dim}(V^*)$, it follows that $\tau$ is surjective (onto) and so **$\tau$ is an isomorphism from $V$ to $V^*$**

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**Theorem** (The Riesz representation theorem)
Let $V$ be a finite-dimensional **nonsingular** metric vector space. The linear functional $\tau:V\rightarrow V^*$ defined by $$\tau(x)=\phi_x$$ where $\phi_x(v)=\langle v,x\rangle$ for all $v\in V$, is an isomorphism from $V$ to $V^*$. It follows that for each $f\in V^*$ there exists a unique vector $x\in V$ for which $f=\phi_x$, that is, $$f(v)=\langle v,x\rangle$$ for all $v\in V$. $\square$
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We'd like to extend the Riesz representation theorem to the case of **subspaces of a metric vector space**. The Riesz representation theorem applies to nonsingular metric vector spaces. Thus if $S$ is a nonsingular subspace of $V$, the Riesz representation theorem applies to $S$ and so all linear functionals on $S$ have the form of an inner product by a (unique) element of $S$. This is nothing new.

As long as $V$ is nonsingular, even if $S$ is **singular**, we can still say something very useful, The reason is that any linear functional $f\in S^*$ can be extended to a linear funtional $g$ in $V^*$ (perhaps in many ways) and since $V$ is nonsingular, the extension $g$ has the form of inner product by a vector in $V$, that is, $$g(v)=\langle v,x\rangle$$ for some $x\in V$. Hence, $f$ also has this form, where its "Riesz vector" is an element if $V$, not necessarily $S$.

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**Theorem** Let $V$ be a metric vector space and let $S$ be a subspace of $V$. If either $V$ or $S$ is nonsingular, the linear transformation $\tau:V\rightarrow S^*$ defined by $$\tau(x)=\phi_x|_S$$ where $\phi_x(v)=\langle v,x\rangle$, is surjective (onto). Hence, for any linear funtional $f\in S^*$ there is a (not necessarily unique) vector $x\in V$ for which $f(s)=\langle s,x\rangle$. Moreover, if $S$ is nonsingular then $x$ can be taken from $S$, in which case it is unique. $\square$
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## Orthogonal Complements and Orthogonal Direct Sums

[[Direct Sums](/s/B1TPajdyH#Direct-Sums)][[Orthogonal Direct Sums](/s/Skk1Z4XJS#Orthogonal-Direct-Sums)][[Wiki: Rank-nullity Theorem](https://www.wikiwand.com/en/Rank%E2%80%93nullity_theorem)]

If $S$ is a subspace of a real inner product space, then the projection theorem says that $$V=S\odot S^{\perp}$$ Hence, the term orthogonal *complement* is justified.

However, in general metric vector spaces, an orthogonal complement may not be a vector space complement. In fact, the case of $S^{\perp}=S$ might be happened if S is totally degenerate in some metric.

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**Definition** A metric vector space $V$ is the orthogonal direct sum of the subspaces $S$ and $T$, written $$V=S\odot T$$ if $V=S\oplus T$ and $S\perp T$. $\Box$
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**Theorem** Let $V$ be a metric vector space. Then $$V=\text{rad}(V)\odot S$$ where $S$ is nonsingular subspace of $V$ and $\text{rad}(V)$ is totally singular.
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**Proof** All subspaces in vector space $V$ including $\text{rad}(V)$ has a complement subspace $S$ such that $V=\text{rad}(V)\oplus S$. But by definition, $\text{rad}(V)\perp S$, so $V=\text{rad}(V)\odot S$. To see that $S$ is nonsingular, if $x\in\text{rad}(S)$ then $x\perp S$ (and $x\perp \text{rad}(V)$ since $x\in V$) and so $x\perp V$, which implies that $x\in\text{rad}(V)\cap S = \{0\}$ by the definition of orthogonal direct sums. Hence, $\text{rad}(S)=\{0\}$ and $S$ is nonsingular. $\square$

這個定理給了我們一個關於任意的 metric vector space $V$ 很好的性質：必定存在一個 nonsingular subspace $S\subseteq V$

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**Theorem** Let $V$ be a nonsingular metric vector space and let $S$ be any subspace of $V$, then
1. $\text{dim}(S) + \text{dim}(S^{\perp}) = \text{dim}(V)$
2. if $V=S+S^{\perp}$, then $V=S\odot S^{\perp}$
3. $S^{\perp\perp}=S$
4. $\text{rad}(S)=\text{rad}(S^{\perp})$
5. $S$ is nonsingular $\Leftrightarrow$ $S^{\perp}$ is nonsingular
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**Proof**
1. the map $\tau:V\rightarrow S^*$ of the last theorem in [Linear Functionals](#Linear-Functionals) is surjective and $$\text{ker}(\tau)=\{x\in V|\phi_x|_S=0\}=\{x\in V|\langle s,x \rangle=0 ,\forall s\in S\}=S^{\perp}$$ Thus, the rank-plus-nullity theorem implies that $$\text{dim}(S^*) + \text{dim}(S^{\perp}) = \text{dim}(V)$$ However, $\text{dim}(S^*)=\text{dim}(S)$ and so part 1 follows.
2. part 1 implies that $$\text{dim}(V) = \text{dim}(S+S^{\perp})\\=\text{dim}(S)+\text{dim}(S^{\perp})-\text{dim}(S\cap S^{\perp})\\=\text{dim}(V)-\text{dim}(S\cap S^{\perp})$$ and so $S\cap S^{\perp}=\{0\}$.
3. part 1 implies that $$\text{dim}(S)+\text{dim}(S^{\perp})=\text{dim}(V)$$ and $$\text{dim}(S^{\perp})+\text{dim}(S^{\perp\perp})=\text{dim}(V)$$ and so $\text{dim}(S^{\perp\perp})=\text{dim}(S)$
4. we have $$\text{rad}(S)=S\cap S^{\perp}=S^{\perp}\cap S^{\perp\perp}=\text{rad}(S^{\perp})$$
5. follows from part 4 $\Box$

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**Theorem** Let S be a subspace of a finite-dimensional metric space $V$. Then $$V=S\odot S^{\perp}$$ if and only if $S$ is nonsingular, that is, if and only if $S\cap S^{\perp}=\{0\}$.
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**Proof.** If $V=S\odot S^{\perp}$ then by definition of orthogonal direct sum, we have $$\text{rad}(S)=S\cap S^{\perp}=\{0\}$$ and so $S$ is nonsingular. Conversely, if $S$ is nonsingular, then $S\cap S^{\perp}=\{0\}$ and so $S\odot S^{\perp}$ exists. Now, the same proof used in part 1 of previous theorem works if $S$ is nonsingular (even if $V$ is singular). To wit, the map $\tau:V\rightarrow S^*$ of the last theorem in [Linear Functionals](#Linear-Functionals) is surjective and $$\text{ker}(\tau)=\{x\in V|\phi_x|_S=0\}=\{x\in V|\langle s,x \rangle=0 ,\forall s\in S\}=S^{\perp}$$ Thus, the rank-plus-nullity theorem gives $$\text{dim}(S^*) + \text{dim}(S^{\perp}) = \text{dim}(V)$$ But $\text{dim}(S^*)=\text{dim}(S)$ and so $$\text{dim}(S\odot S^{\perp})=\text{dim}(S)+\text{dim}(S^{\perp})=\text{dim}(V)$$ It follows that $V=S\odot S^{\perp}$.

## Isometries
[[Wiki: Isometry](https://www.wikiwand.com/en/Isometry)][[Youtube: 台大物理高涌泉相對論](https://youtu.be/vrOvZVQuzIs?t=426)]

We now turn to a discussion of ==structure-preserving== maps on metric vector spaces.

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**Definition** Let $V$ and $W$ be metric vector spaces. We use the same notation $\langle,\rangle$ for the bilinear form on each space. A bijective linear map $\tau:V\rightarrow W$ is called an **isometry** if $$\langle \tau u, \tau v\rangle = \langle u, v \rangle$$ for all vectors $u$ and $v$ in $V$. If an isometry exists from $V$ to $W$, we say that $V$ and $W$ are **isometric** and write $V\approx W$.

It is evident that the set of all isometries from $V$ to $V$ forms a group under composition.

- If $V$ is a nonsingular orthogonal geometry, an isometry of $V$ is called an **orthogonal transformation**. The set $\mathcal{O}(V)$ of all orthogonal transformations on $V$ is a group under composition, known as the **orthogonal group** of $V$.

- If $V$ is a nonsingular symplectic geometry. an isometry of $V$ is called a **symplectic transformation**. The set $\text{Sp}(V)$ of all symplectic transformations on $V$ is a group under composition, known as the **symplectic group** of $V$. $\Box$
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Here are a few of the basic properties of isometries.
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**Theorem** Let $\tau\in \mathcal{L}(V,W)$ be a linear transformation between finite-dimensional metric vector spaces $V$ and $W$.
1. Let $\mathcal{B}=\{v_1, \ldots, v_n\}$ be a basis for $V$. Then $\tau$ is an isometry if and only if $\tau$ is bijective and $$\langle \tau v_i, \tau v_j\rangle=\langle v_i, v_j\rangle$$ for all $i,j$.
2. If $V$ is orthogonal and $\text{char}(F)\neq2$ then $\tau$ is an isometry if and only if it is bijective and $$\langle\tau(v),\tau(v)\rangle=\langle v, v\rangle$$ for all $v\in V$.
3. Suppose that $\tau$ is an isometry and $V=S\odot S^{\perp}$ and $W=T\odot T^{\perp}$. If $\tau(S)=T$ then $\tau(S^{\perp})=T^{\perp}$. In particular, if $\tau\in\mathcal{L}(V)$ is an isometry and $V=S\odot S^{\perp}$ then if $S$ is [$\tau$-invariant](https://hackmd.io/@howayi/HJ9S6QeWS#Invariant-Subspace-and-Reducing-Pairs), so is $S^{\perp}$.
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**Proof** We prove part 3. only. To see that $\tau(S^{\perp})=T^{\perp}$, if $x\in S^{\perp}$ and $t\in T$ then since $T=\tau(S)$, we can write $t=\tau(s)$ for some $s\in S$ and so $$\langle\tau(z),t\rangle = \langle\tau(z),\tau(s)\rangle=\langle z, s\rangle=0$$ whence $\tau(S^{\perp})\subseteq T^{\perp}$. But since $\text{dim}(\tau(S^{\perp}))=\text{dim}(T^{\perp})$, we deduce that $\tau(S^{\perp})=T^{\perp}$. $\square$

## Hyperbolic Spaces
[[Wiki: Hyperbolic Spaces](https://www.wikiwand.com/en/Hyperbolic_space)][[Symmetric, Skew-Symmetric and Alternate Forms](#Symmetric-Skew-Symmetric-and-Alternate-Forms)]

A special type of two-dimensional metric vector space playsan important role in the structure theory of metric vector spaces.

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**Definition** Let $V$ be a metric vector space. If $u, v\in V$ have the property that
- $\langle u, u\rangle=\langle v, v\rangle=0$
- $\langle u, v\rangle = 1$

the ordered pair $(u,v)$ is called a **hyperbolic pair**. Note that
- $\langle v, u\rangle=1$ if $V$ is an orthogonal geometry
- $\langle v, u\rangle=-1$ if $V$ is an symplectic geometry

In either case, the subspace $H=\text{span}(u, v)$ is called a **hyperbolic plane** and any space of the form $$\mathcal{H}=H_1\odot\ldots\odot H_k$$
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